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Question Number 175247 by rexford last updated on 24/Aug/22

	Answered by Ar Brandon last updated on 24/Aug/22

	$I = \int_{0}^{\frac{π}{2}} \sin^{2} 4 ϑ \cos^{5} 4 ϑ d ϑ$ $= \int_{0}^{\frac{π}{2}} \sin^{2} 4 ϑ {(1 - \sin^{2} 4 ϑ)}^{2} \cos 4 ϑ d ϑ$ $= \int_{0}^{\frac{π}{2}} (\sin^{2} 4 ϑ - {2 \sin}^{4} 4 ϑ + \sin^{6} 4 ϑ) \cos 4 ϑ d ϑ$ $= \frac{1}{4} {[\frac{\sin^{3} 4 ϑ}{3} - \frac{{2 \sin}^{5} 4 ϑ}{5} + \frac{\sin^{7} 4 ϑ}{7}]}_{0}^{\frac{π}{2}} = 0$
	Commented by rexford last updated on 25/Aug/22

	$B u t t h e a n s w e r i s n o t z e r o w h e n t y p e d o n c a l c u l a t o r$
	Commented by Ar Brandon last updated on 25/Aug/22

	$Then maybe you should sell the calculator$ $and have a picnic with the money . Haha!$
	Commented by rexford last updated on 25/Aug/22

	$i s e e . . . l o l$
	Commented by Ar Brandon last updated on 25/Aug/22

	$What your calculator gives you is probably$ $the sum of areas bounded by f (x) and the$ $x - axis which is$ ${[\frac{\sin^{3} 4 ϑ}{3} - \frac{{2 \sin}^{5} 4 ϑ}{5} + \frac{\sin^{7} 4 ϑ}{7}]}_{0}^{\frac{π}{8}}$ $= \frac{1}{3} - \frac{1}{5} + \frac{1}{7} = \frac{29}{105} square units$
	Commented by Ar Brandon last updated on 25/Aug/22

	Commented by rexford last updated on 25/Aug/22

	$I g e t i t, B o s s$ $. . . b u t i s t h e r e a n y a p p r o a c h$ $t o r e l a t e t h e i n t e g r a l w i t h t h e$ $b e t a i n t e g r a l i . e 2 \int_{0}^{\frac{Π}{2}} {s i n}^{2 m - 1} θ {c o s}^{2 n + 1} θ d θ$
	Commented by Ar Brandon last updated on 25/Aug/22

	$Yah sure!$ $I = \int_{0}^{\frac{π}{2}} \sin^{2} 4 ϑ \cos^{5} 4 ϑ d ϑ = \frac{1}{4} \int_{0}^{2 π} \sin^{2} t \cos^{5} t d t$ $4 I = \underset{A}{\underset{⏟}{\int_{0}^{\frac{π}{2}} \sin^{2} t \cos^{5} t d t}} + \underset{B}{\underset{⏟}{\int_{\frac{π}{2}}^{π} \sin^{2} t \cos^{5} t d t}} + \underset{C}{\underset{⏟}{\int_{π}^{\frac{3 π}{2}} \sin^{2} t \cos^{5} t d t}} + \underset{D}{\underset{⏟}{\int_{\frac{3 π}{2}}^{2 π} \sin^{2} t \cos^{5} t d t}}$ $For B let t = π - u \Rightarrow B = \int_{0}^{\frac{π}{2}} \sin^{2} (π - u) \cos^{5} (π - u) d u$ $\Rightarrow B = - \int_{0}^{\frac{π}{2}} \sin^{2} u \cos^{5} u d u = - A since \sin (π - u) = \sin u, \cos (π - u) = - \cos u$ $Similarly for C we let t = \frac{3 π}{2} - u and we have$ $- \int_{0}^{\frac{π}{2}} \cos^{2} u \sin^{5} u d u, since \sin (\frac{3 π}{2} - u) = - \cos u and \cos (\frac{3 π}{2} - u) = - \sin u$ $which is equal to - \int_{0}^{\frac{π}{2}} \sin^{2} u \cos^{5} u d u since \int_{0}^{\frac{π}{2}} \sin^{m} u \cos^{n} u d u = \int_{0}^{\frac{π}{2}} \cos^{m} u \sin^{n} u d u$ $And for D when we let t = 2 π - t we have \int_{0}^{\frac{π}{2}} \sin^{2} u \cos^{5} u d u$ $So$ $I = \int_{0}^{\frac{π}{2}} \sin^{2} t \cos^{5} t d t - \int_{0}^{\frac{π}{2}} \sin^{2} u \cos^{5} u d u - \int_{0}^{\frac{π}{2}} \sin^{2} u \cos^{5} u d u + \int_{0}^{\frac{π}{2}} \sin^{2} u \cos^{5} u d u$ $Now you can apply your beta function since we now have the usual limits$ $[0, \frac{π}{2}] . Although that will be superfluous since we can already notice$ $all the partial integrals cancel out themselves$ $A - A - A + A = 0$
	Answered by Mathspace last updated on 25/Aug/22

	$Ψ = \int_{0}^{\frac{π}{2}} {s i n}^{2} (4 θ) {c o s}^{5} (4 θ) d θ$ $=_{4 θ = t} \frac{1}{4} \int_{0}^{2 π} {s i n}^{2} t {c o s}^{5} t d t$ $\int_{0}^{2 π} {s i n}^{2} t {c o s}^{5} t d t$ $= \int_{0}^{\frac{π}{2}} (. . .) d t + \int_{\frac{π}{2}}^{π} (. . .) d t + \int_{π}^{\frac{3 π}{2}} (. . .) d t$ $+ \int_{\frac{3 π}{2}}^{2 π} (. . .) d t$ $\int_{\frac{π}{2}}^{π} {s i n}^{2} t {c o s}^{5} t d t$ $=_{t = \frac{π}{2} + x} - \int_{0}^{\frac{π}{2}} {c o s}^{2} x {s i n}^{5} x d x$ $\int_{π}^{\frac{3 π}{2}} {s i n}^{2} t {c o s}^{5} t d t$ $=_{t = π + x} - \int_{0}^{\frac{π}{2}} {s i n}^{2} x {c o s}^{5} x d x$ $\int_{\frac{3 π}{2}}^{2 π} {s i n}^{2} t {c o s}^{5} t d t$ $=_{t = \frac{3 π}{2} + x} \int_{0}^{\frac{π}{2}} {c o s}^{2} x {s i n}^{5} x d x \Rightarrow$ $4 Ψ = \int_{0}^{\frac{π}{2}} {s i n}^{2} x {c o s}^{5} x d x - \int_{0}^{\frac{π}{2}} {c o s}^{2} x {s i n}^{5} x d x$ $- \int_{0}^{\frac{π}{2}} {s i n}^{2} x {c o s}^{5} x d x + \int_{0}^{\frac{π}{2}} {c o s}^{2} x {s i n}^{5} x d x$ $= 0 \Rightarrow Ψ = 0$
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